Problem: You have found the following ages (in years) of all 5 turtles at your local zoo: $ 30,\enspace 110,\enspace 21,\enspace 87,\enspace 101$ What is the average age of the turtles at your zoo? What is the standard deviation? You may round your answers to the nearest tenth.
Answer: Because we have data for all 5 turtles at the zoo, we are able to calculate the population mean $({\mu})$ and population standard deviation $({\sigma})$ To find the population mean , add up the values of all $5$ ages and divide by $5$ $ {\mu} = \dfrac{\sum\limits_{i=1}^{{N}} x_i}{{N}} = \dfrac{\sum\limits_{i=1}^{{5}} x_i}{{5}} $ $ {\mu} = \dfrac{30 + 110 + 21 + 87 + 101}{{5}} = {69.8\text{ years old}} $ Find the squared deviations from the mean for each turtle. Age $x_i$ Distance from the mean $(x_i - {\mu})$ $(x_i - {\mu})^2$ $30$ years $-39.8$ years $1584.04$ years $^2$ $110$ years $40.2$ years $1616.04$ years $^2$ $21$ years $-48.8$ years $2381.44$ years $^2$ $87$ years $17.2$ years $295.84$ years $^2$ $101$ years $31.2$ years $973.44$ years $^2$ Because we used the population mean $({\mu})$ to compute the squared deviations from the mean , we can find the variance $({\sigma^2})$ , without introducing any bias, by simply averaging the squared deviations from the mean $ {\sigma^2} = \dfrac{\sum\limits_{i=1}^{{N}} (x_i - {\mu})^2}{{N}} $ $ {\sigma^2} = \dfrac{{1584.04} + {1616.04} + {2381.44} + {295.84} + {973.44}} {{5}} $ $ {\sigma^2} = \dfrac{{6850.8}}{{5}} = {1370.16\text{ years}^2} $ As you might guess from the notation, the population standard deviation $({\sigma})$ is found by taking the square root of the population variance $({\sigma^2})$ ${\sigma} = \sqrt{{\sigma^2}}$ $ {\sigma} = \sqrt{{1370.16\text{ years}^2}} = {37\text{ years}} $ The average turtle at the zoo is 69.8 years old. There is a standard deviation of 37 years.